**Appendix A**

In this Appendix A, we provide Lemma A1 and the proofs of Theorem 1, Theorem 2, Theorem 4, and Lemma 1.

**Lemma A1.** *Any solution* **E** *of Problem 2 with l* = 0 *satisfies*

• **n** × **E** = 0 *on* Γ*lif HM3, HB2 and HB3 hold true,*

• **E** = 0 *in D if HM3, HB2 and HM4 hold true; the same result is achieved under hypotheses HM3, HB2 and HM5 or HM3, HB2, HM6 and HM7.*

**Proof.** Consider **E** as the solution of Problem 2 with *l* = 0 and choose **v** = **E** in Equation (6). Since *<sup>a</sup>*(**<sup>E</sup>**, **E**) = 0, we ge<sup>t</sup>

$$0 = \operatorname{Im} (a(\mathbf{E}, \mathbf{E})) = -\int\_{\Omega} \left( \mathbf{E}^\*, \operatorname{curl} \mathbf{E}^\* \right) A\_{\text{ss}} \begin{pmatrix} \mathbf{E} \\ \operatorname{curl} \mathbf{E} \end{pmatrix} + \omega \int\_{\Gamma} \operatorname{Re} (\mathcal{Y}) |\mathbf{n} \times \mathbf{E} \times \mathbf{n}|^2. \tag{A1}$$

Taking account that *ω* > 0, if we assume HM3, HB2, and HB3, we easily get:

$$0 \geq \int\_{\Gamma} \text{Re}(Y) |\mathbf{n} \times \mathbf{E} \times \mathbf{n}|^2 \geq \int\_{\Gamma\_l} \text{Re}(Y) |\mathbf{n} \times \mathbf{E} \times \mathbf{n}|^2 \geq \mathbb{C}\_{Ym} \int\_{\Gamma\_l} |\mathbf{n} \times \mathbf{E} \times \mathbf{n}|^2. \tag{A2}$$

Thus, under the indicated hypotheses, we can conclude that **n** × **E** = 0 on Γ*l*. If HM3, HB2, and HM4 hold true, considering that *ω* > 0, we ge<sup>t</sup>

$$0 \geq -\int\_{\Omega} (\mathbf{E}^\*, \operatorname{curl} \mathbf{E}^\*) A\_{\mathcal{S}^2} \begin{pmatrix} \mathbf{E} \\ \operatorname{curl} \mathbf{E} \end{pmatrix} \geq K\_{dl} \int\_D (|\mathbf{E}|^2 + |\operatorname{curl} \mathbf{E}|^2) \geq K\_{dl} \int\_D |\mathbf{E}|^2 \tag{A3}$$

and we conclude that **E** = 0 in *D*.

> The same result easily follows if we assume HM3, HB2 and HM5, since

$$0 \geq -\int\_{\Omega} (\mathbf{E}^\*, \operatorname{curl} \mathbf{E}^\*) A\_{\otimes \mathbf{t}} \begin{pmatrix} \mathbf{E} \\ \operatorname{curl} \mathbf{E} \end{pmatrix} \geq K\_{\operatorname{cl}} \int\_D |\mathbf{E}|^2,\tag{A4}$$

or if HM3, HB2, HM6 and HM7 hold true, since

$$0 \geq -\int\_{\Omega} (\mathbf{E}^\*, \mathbf{curl} \, \mathbf{E}^\*) A\_{ss} \begin{pmatrix} \mathbf{E} \\ \mathbf{curl} \, \mathbf{E} \end{pmatrix} \geq K\_{ml} \int\_D |\mathbf{curl} \, \mathbf{E}|^2 \tag{A5}$$

and taking account of (5)2 with **J***m* = 0, (1)2 with *M* = 0, (5)1 with **J***e* = 0 and (1)1 with *L* = 0.

**Proof of Theorem 1.** By Lemma A1, there is either a subdomain *D* where the electric field **E** = 0 or a part of the boundary, Γ*l*, where **n** × **E** = 0. We prove that the fields **E**, **B**, **H** and **D** are identically zero in Ω*i*, for all *i* ∈ *I*, if one of the following is true:


*Symmetry* **2020**, *12*, 218

In both cases, we introduce a sufficiently small open ball *B* ⊂ R<sup>3</sup> centered on a point of Ω*i* ∩ Ω*k* or on a point of *∂*Ω*i* ∩ Γ*l*.

In both cases, we have **n** × **E** = 0 on *B* ∩ *∂*Ω*i*. Then, considering the homogeneous version of (5), by (5)3 we ge<sup>t</sup> **n** × **H** = 0 on *B* ∩ *∂*Ω*i*. Then, Equations (5)1 and (5)2 respectively imply that the normal components **n** · **D** = 0 and **n** · **B** = 0 on *B* ∩ *∂*Ω*i*.

Now, we can extend in an analytic way, from Ω*i* to *B* \ Ω*i*, all the components of *κ*, *χ*, *γ* and *ν*. This is possible because of HM8. In both cases, we have to consider that **E**, **B**, **H** and **D** are either trivial fields in *B* \ Ω*i* (in the first of the two cases) or can be trivially extended to *B* \ Ω*i* (in the second of the two cases).

Next, we can show that the fields in *B* are analytic in Ω*i*, ∀*i* ∈ *I*. As a matter of fact:



Thus, by Theorems 6.4 and 7.3 of [22], we can conclude that the electromagnetic fields in *B* are analytic. Since they are equal to zero or can be extended to zero fields in *B* \ Ω*i*, we ge<sup>t</sup> **E** = 0, **B** = 0, **H** = 0, and **D** = 0 in *B*. Once the fields are proved to be equal to zero in *B*, we easily see that they are zero in Ω*i* by the analyticity of the indicated fields in Ω*i*. This procedure can be successively applied to all subdomains allowing us to conclude that the homogeneous version of Problem 2 has only trivial solutions and hence Problem 2 admits a unique solution.

**Proof of Theorem 2.** The homogeneous variational problem defined in the statement of the theorem is similar in form to the homogeneous version of the original problem. The only differences are the reversed roles of **u** and **v** and the change in sign of the imaginary part. Hence, the same proof will also work here. In particular, in the proof of Lemma A1, we can use the fact that (*a*(**<sup>E</sup>**, **E**))<sup>∗</sup> = 0 implies *Im* (*a*(**<sup>E</sup>**, **E**)) = 0, which in turn implies (A1) and hence the conclusions of Lemma A1 hold also for the homogeneous variational problem defined in the statement of the theorem. The arguments for showing the unique continuation results are not affected by the sign of imaginary part of the sesquilinear form. Hence, we can conclude that **v** = 0 is the only solution.

**Proof of Theorem 4.** We prove Theorem 4 by contradiction, as we did in [10]. Due to the similarities with the corresponding proof presented in [10], we report here the main ideas.

As in [10], we ge<sup>t</sup> the result by contradiction and, thus, we assume that:

$$\exists \{\mathbf{u}\_n\}, \ \mathbf{u}\_n \in \mathcal{U} \text{ and } \|\mathbf{u}\_n\|\_{\mathcal{U}} = 1 \,\forall n \in \mathbb{N}, \text{ such that } \lim\_{n \to \infty} \sup\_{\|\mathbf{v}\|\_{\mathcal{U}} \le 1} |a(\mathbf{u}\_n, \mathbf{v})| = 0. \tag{A6}$$

For the space *U*, under hypotheses HD1–HD3, HM1–HM2, and HM13, the following Helmholtz decomposition holds true [24] (p. 86):

$$
\mathcal{U}I = \mathcal{U}\_0 \oplus \mathcal{U}\_1. \tag{A7}
$$

where

$$\mathcal{U}l0 = \{ \mathbf{u} \in \mathcal{U} \mid \mathbf{curl} \,\mathbf{u} = 0 \text{ in } \Omega \text{ and } \mathbf{u} \times \mathbf{n} = 0 \text{ on } \Gamma \}\tag{A8}$$

and

$$\mathcal{U}I\_1 = \{ \mathbf{u} \in \mathcal{U} \mid (P\mathbf{u}, \mathbf{v})\_{0, \Omega} = 0 \,\,\forall \mathbf{v} \in \mathcal{U}\_0 \}. \tag{A9}$$

Thus, for any element of the sequence satisfying (A6), we ge<sup>t</sup>

$$\mathbf{u}\_{n} = \mathbf{u}\_{n0} + \mathbf{u}\_{n1} \tag{A10}$$

with **u***n*0 ∈ *U*0 and **u***n*1 ∈ *U*1.

> Under the assumed hypotheses, one easily gets:

$$\|\|\mathbf{u}\_{n0}\|\|\_{0,\Omega} = \|\mathbf{u}\_{n0}\|\|\_{L} \le \frac{\mathsf{C}\_{PL}}{\mathsf{C}\_{PS}} \|\mathbf{u}\_{n}\|\|\_{0,\Omega} \le \frac{\mathsf{C}\_{PL}}{\mathsf{C}\_{PS}} \|\mathbf{u}\_{n}\|\|\_{L^{\prime}}\tag{A11}$$

$$\|\|\mathbf{u}\_{n1}\|\|\_{\mathcal{U}} \le \frac{\mathcal{C}\_{PS} + \mathcal{C}\_{PL}}{\mathcal{C}\_{PS}} \|\|\mathbf{u}\_{\mathcal{U}}\|\|\_{\mathcal{U}} \tag{A12}$$

$$\lim\_{n \to \infty} \|\mathbf{n} \times \mathbf{u}\_n \times \mathbf{n}\|\_{0, \Gamma} = 0. \tag{A13}$$

Thus, the two sequences, {**<sup>u</sup>***n*<sup>0</sup>} and {**<sup>u</sup>***n*<sup>1</sup>}, subsequences of the sequence satisfying (A6), are bounded in *U*. Since our hypotheses guarantees that uniqueness also holds, then, on a common subsequence of indices, both {**<sup>u</sup>***n*<sup>0</sup>} and {**<sup>u</sup>***n*<sup>1</sup>} weakly converge to zero in *U* and, by the compact embedding of *U*1 in (*L*<sup>2</sup>(Ω))3, which holds true under hypothesis HM13, from {**<sup>u</sup>***n*<sup>1</sup>}, we can extract a subsequence which converges strongly in (*L*<sup>2</sup>(Ω))<sup>3</sup> to **u**ˆ 1. Finally, since both weak convergence in *U* and strong convergence in (*L*<sup>2</sup>(Ω))<sup>3</sup> imply weak convergence in (*L*<sup>2</sup>(Ω))<sup>3</sup> to the same limit, we immediately deduce **u**ˆ 1 = 0.

By setting **u** = **u***n* and **v** = **u***n*0 for any *n*, we ge<sup>t</sup> from the very definition of the sesquilinear form *a*:

$$\|\mathbf{C}\_{PS}\| \|\mathbf{u}\_{n0}\|\_{0,\Omega}^2 \le \frac{c\_0}{\omega^2} |a(\mathbf{u}\_n, \mathbf{u}\_{n0})| + \frac{c\_0 C\_L}{\omega} \|\mathbf{curl}\,\mathbf{u}\_n\|\_{0,\Omega} \|\mathbf{u}\_{n0}\|\_{0,\Omega}.\tag{A14}$$

By the same token, by setting **u** = **u***n* and **v** = **u***n*1 for any *n*, we deduce

$$\begin{split} c\_{0}\mathbb{C}\_{\rm QS} \| \operatorname{curl} \mathbf{u}\_{n1} \| \|\_{0,\Omega}^{2} &\leq \\ \quad \| a(\mathbf{u}\_{n\star}\mathbf{u}\_{n1}) \| + \frac{\omega^{2}\mathbb{C}\_{PL}}{c\_{0}} \| \mathbf{u}\_{n} \| \|\_{0,\Omega} \| \mathbf{u}\_{n1} \| \|\_{0,\Omega} + \omega \mathbb{C}\_{M} \| \mathbf{u}\_{n0} \| \|\_{0,\Omega} \| \operatorname{curl} \mathbf{u}\_{n1} \| \|\_{0,\Omega} + \\ \quad + \omega (\mathbb{C}\_{M} + \mathbb{C}\_{L}) \| \mathbf{u}\_{n1} \| \|\_{0,\Omega} \| \operatorname{curl} \mathbf{u}\_{n1} \| \|\_{0,\Omega} + \omega \mathbb{C}\_{YL} \| \mathbf{n} \times \mathbf{u}\_{n1} \times \mathbf{n} \|\_{0,\Gamma}^{2} .\end{split} \tag{A15}$$

Now, taking into account that {**<sup>u</sup>***n*<sup>0</sup>} and {**<sup>u</sup>***n*<sup>1</sup>} are bounded in *U*, **<sup>u</sup>***n*<sup>1</sup> 0,Ω → 0 on a subsequence, (**n** × **u***n*1 × **n**) 0,Γ → 0, by using inequalities (A14) and (A15), we deduce that we cannot find a subsequence such that either {**<sup>u</sup>***n*<sup>0</sup>} or {curl **<sup>u</sup>***n*<sup>1</sup>} converges to zero in (*L*<sup>2</sup>(Ω))3. As a matter of fact, if one of them did converge to zero in (*L*<sup>2</sup>(Ω))3, then both should do and we would obtain that {**<sup>u</sup>***n*} should converge to zero in *U* against the hypothesis.

Then, we can find a subsequence giving **<sup>u</sup>***n*<sup>1</sup> 0,Ω → 0 and **<sup>u</sup>***n*<sup>0</sup> 0,Ω ≥  > 0. On this subsequence, from inequality (A14), we ge<sup>t</sup>

$$\|\mathbf{u}\_{n0}\|\_{0,\Omega} \le \frac{c\_0}{\omega^2 \mathbf{C}\_{PS}} |a(\mathbf{u}\_{n\prime} \frac{\mathbf{u}\_{n0}}{\|\mathbf{u}\_{n0}\|\_{0,\Omega}})| + \frac{c\_0 \mathbf{C}\_L}{\omega \mathbf{C}\_{PS}} \|\mathbf{curl}\,\mathbf{u}\_{n1}\|\_{0,\Omega}.\tag{A16}$$

By substituting the right-hand side of (A16) for **<sup>u</sup>***n*<sup>0</sup> 0,Ω in inequality (A15), we deduce

$$\begin{split} \left| c\_{0} \left( \mathbf{C}\_{QS} - \frac{\mathbf{C}\_{L} \mathbf{C}\_{M}}{\mathbf{C}\_{PS}} \right) \right| \left| \operatorname{curl} \mathbf{u}\_{n1} \right| \left| \mathbf{l}\_{0,\Omega}^{2} \right| &\leq \\ \frac{\mathbf{C}\_{PS} + \mathbf{C}\_{PL}}{\mathbf{C}\_{PS}} \left| a \left( \mathbf{u}\_{n\prime} \frac{\mathbf{C}\_{PS}}{\mathbf{C}\_{PS} + \mathbf{C}\_{PL}} \mathbf{u}\_{n1} \right) \right| + \frac{\omega^{2} \mathbf{C}\_{PL}}{c\_{0}} ||\mathbf{u}\_{n}||\_{0,\Omega} \|\mathbf{u}\_{n1}\|\_{0,\Omega} + \\ &+ \omega (\mathbf{C}\_{M} + \mathbf{C}\_{L}) ||\mathbf{u}\_{n1}||\_{0,\Omega} ||\operatorname{curl} \mathbf{u}\_{n1}||\_{0,\Omega} + \omega \mathbf{C}\_{NL} ||(\mathbf{n} \times \mathbf{u}\_{n1} \times \mathbf{n})||\_{0,\Gamma}^{2} + \\ &+ \frac{c\_{0} \mathbf{C}\_{M}}{\omega \mathbf{C}\_{PS}} \left| a \left( \mathbf{u}\_{n\prime} \frac{\mathbf{u}\_{n0}}{||\mathbf{u}\_{n0}||\_{0,\Omega}} \right) \right| \left| \operatorname{curl} \mathbf{u}\_{n1} \right| \|\_{0,\Omega} . \end{split} \tag{A17}$$

The right-hand side of inequality (A17) converges to zero on the indicated subsequence and, by hypothesis HM15, we ge<sup>t</sup> curl **<sup>u</sup>***n*<sup>1</sup> 0,Ω → 0, which is against the starting hypothesis.

**Proof of Lemma 1.** We have to analyse just the case when Ω*el* is neither the whole Ω nor the empty set. For all **u** ∈ (*L*<sup>2</sup>(Ω))3, we have

$$\begin{split} |(P\mathbf{u}, \mathbf{u})\_{0, \Omega}|^{2} &= \left| \int\_{\Omega} \mathbf{u}^{\*} P\_{s} \mathbf{u} - j \int\_{\Omega} \mathbf{u}^{\*} P\_{s\mathbf{s}} \mathbf{u} \right|^{2} = \\ &= \left( \int\_{\Omega} \mathbf{u}^{\*} P\_{s} \mathbf{u} \right)^{2} + \left( \int\_{\Omega} \mathbf{u}^{\*} P\_{\partial s} \mathbf{u} \right)^{2} = \\ &= \left( \int\_{\Omega \backslash \Omega\_{d}} \mathbf{u}^{\*} P\_{\mathbb{S}} \mathbf{u} - \int\_{\Omega\_{d}} -\mathbf{u}^{\*} P\_{\mathbb{S}} \mathbf{u} \right)^{2} + \left( \int\_{\Omega\_{d}} \mathbf{u}^{\*} P\_{\mathbb{S}^{\varepsilon}} \mathbf{u} + \int\_{\Omega \backslash \Omega\_{d}} \mathbf{u}^{\*} P\_{\mathbb{S}^{\varepsilon}} \mathbf{u} \right)^{2}. \end{split} \tag{A18}$$

Under assumption HM3, by using Lemma B.1 of [9] with *K*1 = *K*2 = 0, we ge<sup>t</sup> that *Pss* is positive semi definite in Ω*i*, ∀*i* ∈ *I*. Moreover, since Ω*el* is the union of the subdomains Ω*i* of Ω where *Pss* is uniformly positive definite, we ge<sup>t</sup>

$$|(P\mathbf{u},\mathbf{u})\_{0,\Omega}|^2 \ge \left(\int\_{\Omega \, |\, \Omega\_{\mathrm{cl}}|} \mathbf{u}^\* P\_s \mathbf{u} - \int\_{\Omega\_{\mathrm{cl}}} -\mathbf{u}^\* P\_s \mathbf{u} \right)^2 + \mathbb{C}\_1^2 ||\mathbf{u}||\_{0,\Omega\_{\mathrm{cl}}}^4. \tag{A19}$$

However, for all *a*, *b* ∈ R, for any *α* > 0, we have

$$(a-b)^2 \ge (1-a)a^2 + (1-\frac{1}{a})b^2. \tag{A20}$$

Then, using the above inequality for the first addend of the right-hand side of Equation (A19), we ge<sup>t</sup>

$$\|(P\mathbf{u},\mathbf{u})\_{0,\Omega}\|^2 \ge (1-a)\left(\int\_{\Omega\backslash\Omega\_{dl}} \mathbf{u}^\* P\_s \mathbf{u} \right)^2 + (1-\frac{1}{a})\left(\int\_{\Omega\_{dl}} \mathbf{u}^\* P\_s \mathbf{u} \right)^2 + \mathcal{C}\_1^2 ||\mathbf{u}||\_{0,\Omega\_{dl}}^4. \tag{A21}$$

The validity of assumption HM2 guarantees that inequality (28) holds true. Then, by taking account that 1 − 1*α* < 0 for all *α* ∈ (0, <sup>1</sup>), we ge<sup>t</sup>

$$|(P\mathbf{u},\mathbf{u})\_{0,\Omega}| \ge (1-\alpha)\left(\int\_{\Omega\backslash\Omega\_{cl}} \mathbf{u}^\* P\_{\mathbb{S}} \mathbf{u}\right)^2 + \left(\mathbb{C}\_1^2 + (1-\frac{1}{\alpha})\mathbb{C}\_3^2\right)||\mathbf{u}||\_{0,\Omega\_{cl}}^4.\tag{A22}$$

By using (25), we then deduce

$$|(P\mathbf{u},\mathbf{u})\_{0,\Omega}|^2 \ge (1-\mathfrak{a})\mathsf{C}\_5^2||\mathbf{u}||\_{0,\Omega/\Omega\_{cl}}^4 + (\mathsf{C}\_1^2 + (1-\frac{1}{\mathfrak{a}})\mathsf{C}\_3^2)||\mathbf{u}||\_{0,\Omega\_{cl}}^4. \tag{A23}$$

By defining 1 > *α* > *C*23 *<sup>C</sup>*21+*<sup>C</sup>*23 > 0, we have that both terms in (A23) are positive. As a matter of fact, we can think of the right-hand side of (A23) as *s*2 + *t*2, *s*, *t* ∈ R, and, since *s*2 + *t*2 ≥ (*s*+*<sup>t</sup>*)<sup>2</sup> 2 , we ge<sup>t</sup>

$$\begin{split} |(P\mathbf{u},\mathbf{u})\_{0,\Omega}|^{2} &\geq \frac{1}{2} \left( \sqrt{(1-\alpha)}\,\mathrm{C}\_{5} ||\mathbf{u}||\_{0,\Omega\backslash\Omega\_{d}}^{2} + \sqrt{\mathsf{C}\_{1}^{2} + (1-\frac{1}{\alpha})\mathsf{C}\_{3}^{2}} ||\mathbf{u}||\_{0,\Omega\_{d}}^{2} \right)^{2} \geq \\ &\geq \frac{1}{2} \left( \min\left(\sqrt{(1-\alpha)}\,\mathrm{C}\_{5}, \sqrt{\mathsf{C}\_{1}^{2} + (1-\frac{1}{\alpha})\mathsf{C}\_{3}^{2}}\right) \right)^{2} \left( ||\mathbf{u}||\_{0,\Omega\backslash\Omega\_{d}}^{2} + ||\mathbf{u}||\_{0,\Omega\_{d}}^{2} \right)^{2} = \\ &= \frac{1}{2} \min\left( (1-\alpha)\mathsf{C}\_{5}^{2}, \mathrm{C}\_{1}^{2} + (1-\frac{1}{\alpha})\mathsf{C}\_{3}^{2} \right) ||\mathbf{u}||\_{0,\Omega}^{4} .\end{split} \tag{A24}$$
